Wikipedia:Reference desk/Archives/Mathematics/2009 November 20

= November 20 =

Counting system
StuRat (talk) 18:27, 20 November 2009 (UTC)

What's the name of the common counting system where you draw 1-4 lines and then put a slash through the four lines when you get to 5 ? It looks something like this:

||||/ ||||   |||   ||   |   |||/   ||||   |||   ||   |  ||/|=5 ||||=4 |||=3 ||=2 |=1  |/||   ||||   |||   ||   |  /|||   ||||   |||   ||   | /||||   ||||   |||   ||   |

StuRat (talk) 10:15, 20 November 2009 (UTC)


 * Tally. -- PS T  10:39, 20 November 2009 (UTC)


 * Thanks. StuRat (talk) 18:25, 20 November 2009 (UTC)


 * Also five-barred gate, rather more specific than tally.→86.148.187.62 (talk) 21:03, 20 November 2009 (UTC)

Investigation of an integral
I'm working my way through some assorted questions, and stumbled across the following:

$$\mbox{Investigate the integral: }\int_{0}^{3}\frac{1}{(1-x)^2}\, dx$$

I'm afraid I can't really get anywhere with it, and when I finally succombed to Mathematica it told me it didn't converge between 0 and 3. Any suggestions for what I should try? --80.229.152.246 (talk) 22:24, 20 November 2009 (UTC)
 * Without knowing what level you're at, I can't guess what investigations you're supposed to carry out, but the value of the integral is fairly obviously +∞. Algebraist 22:31, 20 November 2009 (UTC)
 * Yeah, I got that much, but didn't really get anything else apart from that. Also, how would you go about setting out a proof of that fact? As for my current level, just before undergraduate. --80.229.152.246 (talk) 23:07, 20 November 2009 (UTC)
 * I would prove it by noting that the function is clearly measurable, that the Lebesgue monotone convergence theorem implies that the integral you want is the limit of $$\int_{0}^{1-1/n}\frac{1}{(1-x)^2}\, dx + \int_{1+1/n}^3\frac{1}{(1-x)^2}\, dx$$, and that the Fundamental theorem of calculus allows you to calculate those integrals and see that the limit is indeed plus infinity. That's not what I'd've written before I was an undergrad, though. Algebraist 23:14, 20 November 2009 (UTC)
 * Thank you for that. While I agree that it's not exactly what a pre-undergrad would write, I'll still look into it. It's sure to be interesting, regardless. --80.229.152.246 (talk) 23:35, 20 November 2009 (UTC)

Since as a pre-undergrad. you are likely to be more familiar with Riemann integral, rather than Lebesgue integral, here is a proof you can follow based on those ideas: Now by choosing h small enough, you can make the lower bound for the integral as large as you want. Hence the original integral evaluates to +∞. QED. Abecedare (talk)
 * 1) Show that for any $$0 \int_{1-h}^{1+h}\frac{1}{(1-x)^2}\, dx$$ (Hint: the integrand is positive)
 * 2) Observe that, $$ \int_{1-h}^{1+h}\frac{1}{(1-x)^2}\, dx \ge \int_{1-h}^{1+h}\frac{1}{h^2}\, dx = \frac{2}{h}$$.


 * Also, if you can do simple linear changes of variable, you may observe that
 * $$\int^{a}_{a/2}\frac{1}{x^2}\, dx = 2\int_{a}^{2a}\frac{1}{x^2}\, dx>0,$$
 * that also makes it clear that the integral diverges.--pma (talk) 05:07, 21 November 2009 (UTC)
 * Or, even simplier, if you haven't any problem with a linear change of variable for an "improper Riemann integral", as it is the integral of 1/x2 over (0,1]:
 * $$\int^1_0\frac{1}{x^2}\, dx = 2\int_0^2\frac{1}{x^2}\, dx>0,$$
 * thus the only possible value of the LHS is $$+\infty$$. --pma (talk) 10:29, 21 November 2009 (UTC)

Thanks for all the help. --80.229.152.246 (talk) 17:26, 21 November 2009 (UTC)